Transforming this hypergeometric function with [1]} gives
\(R=\dfrac{\Gamma (k+n+2)}{\Gamma (k+1)\Gamma (k+2)}-\dfrac{\Gamma (n+1)\Gamma (n+2)}{\Gamma (k)\Gamma (k+1)\Gamma (n+2-k)}\,\mbox{}_{3}F_{2}\!\left(\genfrac{}{}{0.0pt}{}{1-k,-k,1}{n-k+2,2};1\right)\) 
